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A projectile has to be fired to hit a target 128 km away. It is assumed that the effects of friction due to air can be ignored and the projectile is fired on Earth. There is only one force acting on the projectile while it is in motion and that is the gravitational force of attraction of the Earth accelerating it vertically downwards.
If the angle at which the projectile is fired made with the horizontal is X, it has an initial vertical velocity of V*sin X and an initial horizontal velocity of V*cos X. The vertical velocity at a time t is equal to V(t) = V*sin X - gt. The projectile can only travel for a distance during which V(t) is greater than 0. This gives the range of the projectile as V*cos X*(V*sin X)/g
cos X*sin X is maximum when X = 45 degrees and this gives the maximum range of the projectile for a given initial velocity as V^2/g.
For the projectile to travel 128 km, the minimum initial velocity at which it can be launched is determined by solving V^2/g = 128*1000
=> V^2 = 128000/9.8
=> V = 114.28 m/s
The projectile would have to be launched at a minimum velocity of 114.28 m/s to be able to hit the target 128 km away.
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